Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sqrt {{x^2} + 4}  - 2}}{{{x^2}}},khi\,\,\,x \ne 0\\2a - \frac{5}{4}{\rm{&nb

* Hướng dẫn giải

Ta có: \(f\left( 0 \right) = 2a - \frac{5}{4}.\) 

\(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 4}  - 2}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {{x^2} + 4}  - 2} \right)\left( {\sqrt {{x^2} + 4}  + 2} \right)}}{{{x^2}\left( {\sqrt {{x^2} + 4}  + 2} \right)}}\) 

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + 4 - 4}}{{{x^2}\left( {\sqrt {{x^2} + 4}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {{x^2} + 4}  + 2}} = \frac{1}{4}.\) 

Hàm số liên tục tại \(x = 0 \Leftrightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} f\left( x \right) \Leftrightarrow 2x - \frac{5}{4} = \frac{1}{4} \Leftrightarrow a = \frac{3}{4}.\)